How many grams of oxygen are required to produce 100 mols of water H2O in a synthesis reaction with excess hydrogen?
The correct answer and explanation is:
To find out how many grams of oxygen are required to produce 100 moles of water (H₂O), we need to understand the stoichiometry of the synthesis reaction between hydrogen and oxygen:
The balanced chemical equation for the formation of water is:
2H2(g)+O2(g)→2H2O(l)2H_2 (g) + O_2 (g) \rightarrow 2H_2O (l)
From the equation, we can see that 1 mole of O₂ produces 2 moles of H₂O.
Step 1: Moles of Oxygen Needed
If 2 moles of H₂O are produced from 1 mole of O₂, then to produce 100 moles of H₂O, we can use a simple proportion: 1 mole of O₂2 moles of H₂O=x moles of O₂100 moles of H₂O\frac{1 \text{ mole of O₂}}{2 \text{ moles of H₂O}} = \frac{x \text{ moles of O₂}}{100 \text{ moles of H₂O}}
Solving for x: x=100×12=50 moles of O₂x = \frac{100 \times 1}{2} = 50 \text{ moles of O₂}
Thus, 50 moles of oxygen are required to produce 100 moles of water.
Step 2: Mass of Oxygen Required
The molar mass of oxygen (O₂) is approximately 32 grams per mole. To find the mass of 50 moles of oxygen, we multiply the number of moles by the molar mass: Mass of O₂=50 moles×32 g/mol=1600 grams of O₂\text{Mass of O₂} = 50 \text{ moles} \times 32 \text{ g/mol} = 1600 \text{ grams of O₂}
Final Answer:
To produce 100 moles of water, 1600 grams of oxygen are required.
This calculation shows how stoichiometry helps in converting moles of a product into the required amount of reactant. The key is balancing the chemical equation and using the molar relationships between reactants and products.